Circles

(i) The tangent at any point of a circle is perpendicular to the radius through the point of contact.

(ii) A circle may have two parallel tangents

(iii) A tangent to a circle intersects it in one point.

(iv) Secant is a line intersecting a circle in two points

(v) The angle between tangent at a point on a circle and the radius through the point is 90°

A circle can have infinite tangents.

since QT is a tangent to the circle at T and OT is radius,

Therefore OT perpendicular QT

It is given that OQ=25 cm and QT= 24 cm

By Pythagoras theorem we have

Given that O is the center of the circle and OP =17 cm and the radius of the circle OT=8cm.

We need to find the length of the segment PT.

The line PT is the tangent line to the circle at the point T, the line through the centre is perpendicular to PT.

Given: PQ is a tangent to the circle intersect at OP=13cm and OQ=5 cm

Proof: In right triangle OQP

Let the two circle intersect at a point X and Y , XY is the common chord.

Suppose A is a point on their common chord and AM and AN be the tangent drawn from A to the circle

AM is the tangent and AXY is a secant.

AM² = AX×AY ............(i)

AN is the tangent and AXY is the secant.

AN² = AX×AY ............(i)

Therefore, from equations (i) and (ii), we get,

AM = AN.

Given: the sides of a quadrilateral touch a circle

To prove: the sum of a pair of opposite sides is equal to the sum of the other pair.

Proof:

From the theoram which states that the lengths of the two tangents drawn from an external point to a circle are equal

From points A the tangents drawn are AP and AS,

AP = AS .... (1)

From points B the tangents drawn are BP and BQ,

BP = BQ ..... (2)

From points D the tangents drawn are DR and DS,

DR = DS ....(3)

From points C the tangents drawn are CR and CQ,

CR = CQ ..... (4)

Add 1,2,3 and 4 to get

AP+BP+DR+CR = AS+BQ+DS+CQ

(AP+BP)+(DR+CR )= (AS+DS)+ (BQ+CQ)

AB+ DC = AD + BC

Hence proved

Given: AB and Ac are tangent to the circle with centre O

PQ is tangent to the circle at X which intersect AB and Ac in P and Q

To find : Perimeter of triangle APQ

Proof:

Three tangent AB,CD and BD of a circle such as AB and CD are two parallel tangent BD intercept an angle BOD at the centre.

To Prove:

Given: PA and PB are tangent of a circle PA= 10 cm and angle APB= 60^o

Let O be the center of the given circle and C be the point of intersection of OP and AB

In triangle PAC and triangle PBC

PA = PB (tangent from an external point are equal)

APC =BPC (tangent from an external point are equally inclined to the segment joining center to the point)

PC =PC (common)

Given: PA and PB are tangent to the circle with centre O

CD is tangent to the circle at E which intersect PA and PB in C and D

To find : Perimeter of triangle PCD

Proof:

Let ABC be the right angled triangle such that angle B=90^o, BC=6cm, AB= 8cm. Let O be the centre and r be the radius of the in circle.

AB, BC and CA are tangent to the circle at P,N and M

OP=ON=OM=r (radius of the circle)

Given: Two tangent segments PA and PB are drawn to a circle with centre O such that .
To prove: OP = 2 AP
Proof:
Construct the figure according to the conditions given.

Here
In triangle OAP and OBP,
PA = PB (Length of Tangents from external point are equal)

OA = OB (Radii of same circle )

OP = OP (common)

Δ OAP ∼ Δ OBP ( By SSS criterion)

∠OPA = ∠OPB = 60^o.

In Triangle OAP ,
∠OAP = 90^o (By theoram which states that tangent to a circle is perpendicular to the radius through the point of contact)
We know in a right angle triangle

⇒ sin 60^o= AP / OP ,
i.e 1/2 = AP / OP

So,
OP = 2 AP
Hence proved.

Given: If is isosceles with AB = AC and C (O, r) is the incircle of the touching BC at L.

To prove: L bisect BC.

Proof:

Construct the figure according to given condition.




AB = AC (given)

From the theorem which states that the lengths of two tangents drawn from external point to a circle are equal. ..... (1)

As tangents AP and AQ are drawn from the external point A.

AP = AQ

Also,

AB = AC

⇒ AP + PB = AQ + QC

⇒ AP + PB = AP + QC

⇒ PB = QC


From (1) as tangents BP and BL are drawn from external point B,

And tangents CQ and CL are drawn from external point C.

⇒ BP = BL ..... (3)

CQ = CL ...... (4)

As we have proved PB = QC

From 3 and 4

BL = CL

⇒ L bisects BC.

Hence proved.

Given: In the above figure, O is the centre of the circle and BCD is tangent to it at C.
To prove: ∠BAC + ∠ACD = 90°
Proof:

In ΔOAC

OA = OC [radii of same circle]

⇒ ∠OCA = ∠OAC [angles opposite to equal sides are equal]

⇒ ∠OCA = ∠BAC [1]

Also,

OC ⊥ BD [Tangent at any point on a circle is perpendicular to the radius through point of contact]

⇒ ∠OCD = 90°

⇒ ∠OCA + ∠ACD = 90°

⇒ ∠BAC + ∠ACD = 90° [From 1]

Hence Proved

Let us label two circles as 'a' and 'b'



As TQ and TP are tangents to circle a,

And TP and TR are tangents to circle b.

By theorem which states that the lengths of the two tangents drawn from external point to a circle are equal.

AD = DS = 5cm (tangent from an external point)

Since AD=23 cm

So,

AR + RD + AD

AR + 5 = 23 cm

AR = 18 cm ------(i)

and AQ = AR

since AR = 18 cm

So, AQ + QB = AB

Now OP and OQ are radius of the circle. So from tangent P and Q

_{OA=10 cm}

_{As we know Perpendicular from centre to the chord bisects the chord.}

_{So AM=MB=8cm}

_{Using Pythagoras theorem in triangle AOM}

Given : AB, BC and AC are tangents to the circle at E, D and F.
BD = 30 cm and DC = 7 cm and ∠BAC = 90°
Recall that tangents drawn from an exterior point to a circle are equal in length
Hence BE = BD = 30 cm
Also FC = DC = 7 cm
Let AE = AF = x → (1)
Then AB = BE + AE = (30 + x)
AC = AF + FC = (7 + x)
BC = BD + DC = 30 + 7 = 37 cm
Consider right Δ ABC, by Pythagoras theorem we have
BC² = AB² + AC²
⇒ (37)² = (30 + x)² + (7 + x)²
⇒ 1369 = 900 + 60x + x² + 49 + 14x + x²
⇒ 2x² + 74x + 949 – 1369 = 0
⇒ 2x²+ 74x – 420 = 0
⇒ x² + 37x – 210 = 0
⇒ x² + 42x – 5x – 210 = 0
⇒ x (x + 42) – 5 (x + 42) = 0
⇒ (x – 5) (x + 42) = 0
⇒ (x – 5) = 0 or (x + 42) = 0
⇒ x = 5 or x = – 42
⇒ x = 5 [Since x cannot be negative]
∴ AF = 5 cm [From (1)]
Therefore AB =30 +x = 30 + 5 = 35 cm
AC = 7 + x = 7 + 5 = 12 cm
Let ‘O’ be the centre of the circle and ‘r’ the radius of the circle.
Join point O, F; points O, D and points O, E.
From the figure,
Area of (ΔABC) = Area (ΔAOB) + Area (ΔBOC) + Area (ΔAOC)
∴ r = 5
Thus the radius of the circle is 5 cm

Triangle AOP is an isosceles triangle because OA=OP as they are the radius of the circle. We know that radius of the circle is always perpendicular to the tangent at the point of contact.

Here OB is the radius and BC is the tangent and B is the point of contact, Therefore

Given: PQ and RS are the two common tangent to the two circle

To Proof: A is the point of intersection of PQ and RS

We know that , length of two tangent drawn from an exterior point to acirclr are equal.

Therefore

PA = RA----------------- (i)

QA = SA ------------------- (ii)

Adding two equations we get

PA + QA =RA + SA

PQ =RS (proved)

We know that ∠ADO’ = 90° (since O’D is perpendicular to AC)

As we know radius is perpendicular to the tangent.

So, OC ⊥ AC

⇒ ∠ACO = 90°

In ΔADO’ and ΔACO,

∠ADO’ = ∠ACO (each 90°)

∠DAO = ∠CAO (common)

By AA criteria,

ΔADO’ ∼ ΔACO

As we know corresponding sides of a triangle are in ratio.

AO = AO’ + O’X + OX

As radii of two circles are equal.

⇒ AO = AO’ + AO’ + AO’

= 3 AO’

Given that OQ: PQ=3:4

Let ratio coefficient =x, so

OQ=3x and PQ=4x

We know that a tangent to a circle is perpendicular to the radius at the point of tangency

So

Then applying Pythagoras theorem in triangle POQ

In the diagram AB is the chord touching the smaller circle. We have the right angled triangle OO'B

By Pythagoras theorem

Now since the chord of the larger circle which touches the smaller circle is bisected at the point of contact

We have

AB= 2 × 24= 48 cm

So ans is 18 cm.

Let PQ and PR touch the circle at points S and U respectively. Join O with P, Q, R, S and U

We have OS = OT = OU = 6cm

QT = 12 cm and TR = 9cm

QR= QT +TR = 12cm + 9cm = 21 cm

Now QT = QS=12 cm (tangent from the same point)

TR =RU=9cm

Let PS =PU =x cm

Then PQ=PS+SQ- (12 + x)cm and PR = PU+RU= (9+x)cm

It is clear that

Firstly consider that the given circle will touch the given circle will touch the sides AB and AC of the triangle at a point E and F respectively.

Let AF=x

Now in triangle ABC

CF = CD=6cm

(Tangent drawn from an external point to a circle are equal. Here tangent is drawn from external point C)

BE = BD =8cm (Tangent drawn from an external point to a circle are equal. Here tangent is drawn from external point B)

AE = AF =X

Now AB= AE + EB =x + 8

Also BC = BD+ DC = 8+6 =14 and CA= CF+FA = 6+ x

Now we get all side of the triangle and its area can be find by using hero’s formula

As we know that the tangents drawn from an external point to a circle are equal.

Therefore, PQ = PR

Also, from the figure, PQR is an isosceles triangle, because PQ = PR

Therefore, ∠RQP =∠QRP (Because the corresponding angles of the equal sides of the isosceles triangle are equal)
And, from the angle sum property of a triangle,

∠RQP + ∠QRP + ∠RPQ = 180^o

∠RQP + ∠RQP + ∠RPQ = 180^o

2∠RQP +∠RPQ = 180^o

2∠RQP +30^o = 180^o

2∠RQP = 180^o - 30^o

2∠RQP = 150^o

∠RQP = 150^o/2

Therefore, ∠RQP = 75^o

SR || QP and QR is a transversal

∵ ∠SRQ = ∠ PQR …[Alternate interior angle]

∴ ∠SRQ = 75°

⇒ ∠ORP = 90°…[Tangent is Perpendicular to the radius through the point of contact]

∠ORP = ∠ORQ + ∠QRP

⇒ 90° = ∠ORQ + 75°

⇒ ∠ORQ = 15°

Similarly, ∠ RQO = 15°

In Δ QOR,

∠QOR + ∠QRO + ∠OQR = 180°

⇒ ∠QOR + 15° + 15° = 180°

⇒ ∠QOR = 150°

⇒ ∠QSR = ∠QOR/2

⇒ ∠QSR = 150°/2 = 75°

In ΔRSQ,

∠RSQ + ∠QRS + ∠RQS = 180°

⇒ 75° + 75° + ∠RQS = 180°

∠RQS = 30º

Given:

OQ = 12 cm

Property:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

By above property, ∆POQ is right-angled at ∠OPQ.

Therefore,

By Pythagoras Theorem in ∆POQ,

OP² + PQ² =OQ²

⇒ PQ² = OQ² – OP²

⇒ PQ= √( OQ² – OP²)

⇒ PQ= √(12² – 5²)

⇒ PQ= √(144 – 25)

⇒ PQ = √119 cm

Hence, PQ = √119 cm

Given:

PB = 10 cm

CQ = 2 cm

Property:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Using the above property,

PA = PB = 10 cm (tangent from P)

And,

CA = CQ= 10 cm (tangent from C)

Now,

PC = PA – CA

= 10 cm – 2 cm

= 8 cm

Hence, PC = 8 cm

Given:

OQ = 25 cm

PQ = 24 cm

Property:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

By above property, ∆POQ is right-angled at ∠OPQ.

Therefore,

By Pythagoras Theorem in ∆POQ,

OP² + PQ² =OQ²

⇒ OP² = OQ² – PQ ²

⇒ OP= √( OQ² – PQ ²)

⇒ OP= √(25² – 24²)

⇒ OP= √(625 – 576)

⇒ OP = √49 cm

⇒ OP = 7 cm

Hence, OP = 7 cm

Given:

Radius of circle (say PO) = 4 cm

Let AB ∥ CD be two tangents which meets the circle at P and Q respectively. And, O be the center of circle.

Property:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

By above property, we can say that distance between two parallel tangents of a circle is equal to its diameter.

Therefore,

PQ = 2 × PO

= 2 × 4 cm

= 8 cm

Hence, Distance between tangents = 8 cm

Given:

AB (say) = 4 cm

Radius (OB) = 3 cm

Property:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

By above property, ∆AOB is right-angled at ∠ABO.

Therefore,

By Pythagoras Theorem in ∆POQ,

OA² = OB² + BA²

⇒ OA= √(OB² + BA²)

⇒ OA= √(3² + 4²)

⇒ OA= √(9 + 16)

⇒ OA = √25 cm

⇒ OA = 5 cm

Hence, distance of A from center = 5 cm

Given:

OA (say) = 5 cm

AB = 4 cm

Let AC be the tangent which meets the circle at the point B and O be the center of circle.

Property:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

By above property, ∆AOB is right-angled at ∠OBA.

Therefore, by Pythagoras Theorem,

AB² + OB² =AO²

⇒ OB ² = AO² – AB²

⇒ OB= √(AO² – AB²)

⇒ OB= √(5² – 4²)

⇒ OB= √(25 – 16)

⇒ OB = √9

⇒ OB= 3 cm

Hence, Radius = 3 cm

Given:

∠APB = 80°

Property 1:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2:Sum of all angles of a quadrilateral = 360°.

Property 3:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By property 1,

∠PAO = 90°

∠PBO = 90°

By property 2,

∠APB + ∠PAO + ∠PBO + ∠AOB = 360°

⇒ ∠AOB = 360° - ∠APB + ∠PAO + ∠PBO

⇒ ∠AOB = 360° - (80° + 90° + 90°)

⇒ ∠AOB = 360° - 260°

⇒ ∠AOB = 100°

Now, in ∆POA and ∆POB

OA = OB [∵ radius of circle]

PA = PB [By property 3 (tangent from P)]

OP = OP [∵ common]

∴ By SSS congruency,

∆POA ≅ ∆POB

Hence, by CPCTC

∠POA = ∠POB

Now,

∠AOB = 100°

⇒ ∠POA + ∠POB = 100° [∵∠AOB = ∠POA + ∠POB]

⇒ ∠POA + ∠POA = 100° [∵∠POA = ∠POB]

⇒ 2∠POA = 100°

⇒ ∠POA = 50°

Hence, ∠POA = 50°

Given:

∠QTP = 100°

Property 1:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2:Sum of all angles of a quadrilateral = 360°.

By property 1,

∠OPT = 90° and ∠OQT = 90°

And,

By property 2,

∠QTP + ∠OPT + ∠OQT + ∠POQ = 360°

⇒ ∠POQ = 360° – ∠QTP + ∠OPT + ∠OQT

⇒ ∠POQ = 360° – 100° + 90° + 90°

⇒ ∠POQ = 360° – 280°

⇒ ∠POQ = 80°

Hence, ∠POQ = 80°

Given:

Radius of circle (say PO) = 5 cm

Let AB ∥ CD be two tangents which meets the circle at P and Q respectively. And, O be the center of circle.

Property:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

By above property, we can say that distance between two parallel tangents of a circle is equal to its diameter.

Therefore,

PQ = 2 × PO

= 2 × 5 cm

= 10 cm

Hence, Distance between tangents = 10 cm

Given:

∠POQ = 110°

Property 1:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2:Sum of all angles of a quadrilateral = 360°.

By property 1,

∠TPO = 90°

∠TQO = 90°

By property 2,

∠POQ + ∠ TPO + ∠ TQO + ∠PTQ = 360°

⇒ ∠PTQ = 360° - ∠POQ + ∠ TPO + ∠ TQO

⇒ ∠PTQ = 360° - (110° + 90° + 90°)

⇒ ∠PTQ = 360° - 290°

⇒ ∠PTQ = 70°

Hence, ∠PTQ = 70°

Given:

PB = 10 cm

CQ = 2 cm

Property:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Using the above property,

PA = PB = 10 cm (tangent from P)

DB = DQ= 10 cm (tangent from D)

And,

CA = CQ= 10 cm (tangent from C)

Now,

Perimeter of ∆PCD = PC + CD + DP

= PC + CQ + QD + DP

= PC + CA + DB + PD [∵CA = CQ and DB = DQ]

= PA + PB [∵PA = PC + CA and PB = PD + BD]

= 10 cm + 10 cm

= 20 cm

Hence, Perimeter of ∆PCD = 20 cm

Property 1:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2:Sum of all angles of a triangle = 180°.

By property 1, ∆POQ is right-angled at ∠OPQ (i.e., ∠OPQ = 90°).

∵ ∆POQ is an isosceles triangle

∴ ∠POQ = ∠OQP

By property 2,

∠POQ + ∠OQP + ∠QPO = 180°

⇒ ∠POQ + ∠OQP = 180° - ∠QPO

⇒ ∠POQ + ∠OQP = 180° - 90°

⇒ ∠POQ + ∠OQP = 180° - 90°

⇒ ∠POQ + ∠OQP = 90°

⇒ ∠OQP + ∠OQP = 90° [∵∠POQ = ∠OQP]

⇒ 2∠OQP = 90°

⇒ ∠OQP = 45°

Hence, ∠OQP = 45°

Given:

CP = 11 cm

BC = 7 cm

Property:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Using the above property,

CP = CQ = 11 cm (tangent from C)

BQ = BR (tangent from B)

And,

AP = AR (tangent from A)

Now,

BR = BQ = CQ – CB

= 11 cm – 7 cm

= 4 cm

Hence, BR = 4 cm

Property 1:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2:Sum of all angles of a straight line = 180°.

Property 3:Sum of all angles of a triangle = 180°.

By property 1, ∆OAB is right-angled at ∠OAB (i.e., ∠OAB = 90°) and ∆PBA is right-angled at ∠ PBA (i.e., ∠ PBA = 90°)

Clearly,

∠b + ∠c = ∠OAB

⇒ ∠b + ∠c = 90°

⇒ ∠b = 90° - ∠c

Similarly,

∠d + ∠e = ∠PBA

⇒ ∠d + ∠e = 90°

⇒ ∠e = 90° - ∠d

Now,

∠a = ∠b = 90° - ∠c [∵ OA = OC (Radius)]

And,

∠e = ∠f = 90° - ∠d [∵ PB = PC (Radius)]

By property 2,

∠a + ∠f + ∠ACB = 180°

⇒ ∠ACB = 180° – ∠a – ∠f

⇒ ∠ACB = 180° – (90° - ∠c) – (90° - ∠d)

⇒ ∠ACB = 180° – 90° + ∠c – 90° + ∠d

⇒ ∠ACB = ∠c + ∠d

Now, in ∆ACB

By property 3,

∠ACB + ∠c + ∠d = 180°

⇒ ∠ACB + ∠ACB = 180° [∵∠ACB = ∠c + ∠d]

⇒ 2∠ACB = 180°

⇒ ∠ACB = 90°

Hence, ∠ACB = 90°

Given:

AR = 4 cm

BR = 3 cm

AC = 11 cm

Property:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Using the above property,

AR = AQ = 4 cm (tangent from A)

BR= BP (tangent from B)

And,

CP = CQ (tangent from C)

Also,

CQ = CA – AQ = 11 cm – 4 cm = 7 cm

Now,

BC = BP + PC

= BR + CQ [∵ BR = BP and CP = CQ = 7 cm]

= 3 cm + 7 cm

= 10 cm

Hence, BC = 10 cm

Given:

BC = 6 cm

AB = 8 cm

Property 1:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 3:Sum of all angles of a quadrilateral = 360°.

By property 1,

AP = AQ (Tangent from A)

BP = BR (Tangent from B)

CR = CQ (Tangent from C)

∵ ABC is a right-angled triangle, ∴ by Pythagoras Theorem

AC² = AB² + BC²

⇒ AC² = 8² + 6²

⇒ AC² = 64 + 36

⇒ AC² = 100

⇒ AC = √100

⇒ AC = 10 cm

Clearly,

AQ + QC = AC = 10 cm

⇒ AP + RC = 10 cm [∵ AQ = AP and QC = RC]

Also,

AB + BC = 8 cm + 6 cm = 14 cm

⇒ AP + PB + BR + RC = 14 cm [∵ AB = AP + PB and BC = BR + RC]

⇒ AP + RC + PB + BR = 14 cm

⇒ 10 cm + BR + BR = 14 cm [∵ AP + RC = 10 cm and PB = BR]

⇒ 10 cm + 2BR = 14 cm

⇒ 2BR = 14 cm – 10 cm = 4 cm

⇒ BR = 2 cm

Now,

∠BPO = 90° [By property 3]

∠BRO = 90° [By property 3]

∠PBM = 90° [Given]

Now by property 2,

∠BPO + ∠BRO + ∠PBM + ∠ROP = 360°

⇒ ∠ROP = 360° - ∠BPO + ∠BRO + ∠PBM

⇒ ∠ROP = 360° - (90° + 90° + 90°)

⇒ ∠ROP = 360° - 270°

⇒ ∠ROP = 90°

Now, ∵ ∠ROP = 90° and BP = BR which are adjacent sides

∴ Quadrilateral PBRO is a square

⇒ PO = BR = 2 cm

Hence, Radius = 2 cm

Given:

BR = 4 cm

CP = 11 cm

Property:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Using the above property,

BR = BQ = 4 cm (tangent from B)

And,

CP = CQ = 11 cm (tangent from C)

Now,

BC = CQ – BQ

= 11 cm – 4 cm

=7 cm

Hence, BC = 7 cm

Given:

∠POR = 120°

Property 1:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2:Sum of all angles of a straight line = 180°.

Property 3:Sum of all angles of a triangle = 180°.

By property 1,

∠PQO = 90°

By property 2,

∠POQ + ∠POR = 180°

⇒ ∠POQ + 120° = 180°

⇒ ∠POQ = 180° - 120°

⇒ ∠POQ = 60°

Now by property 3 in ∆OPQ,

∠POQ + ∠PQO + ∠OPQ = 180°

⇒ ∠OPQ = 180° - ∠POQ + ∠PQO

⇒ ∠OPQ = 180° - (60° + 90°)

⇒ ∠OPQ = 180° - 150°

⇒ ∠OPQ = 30°

Hence, ∠OPQ = 30°

Given:

AO (say) = CO (say) = 5 cm

BO (say) = 3 cm

Let AC be the tangent which meets the circle at the point B and O be the center of circle.

Property:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

By above property, ∆AOB is right-angled at ∠OBA and ∆COB is right-angled at ∠OBC.

Therefore,

By Pythagoras Theorem in ∆AOB,

AB² + OB² =AO²

⇒ AB ² = AO² – OB ²

⇒ AB= √(AO² – OB ²)

⇒ AB= √(5² – 3²)

⇒ AB= √(25 – 9)

⇒ AB = √16

⇒ AB= 4 cm

Similarly,

By Pythagoras Theorem in ∆COB,

AB² + OB² =CO²

⇒ CB ² = CO² – OB ²

⇒ CB= √(CO² – OB ²)

⇒ CB= √(5² – 3²)

⇒ CB= √(25 – 9)

⇒ CB = √16

⇒ CB= 4 cm

Now,

AC = AB + BC

= 4 cm + 4 cm

= 8 cm

Hence, Length of chord = 8 cm

Property:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By the above property,

AP = AS (tangent from A)

BP = BQ (tangent from B)

CR = CQ (tangent from C)

DR = DS (tangent from D)

Now we add above 4 equations,

AP + BP + CR + DR = AS + BQ + CQ + DS

⇒ AB + CD = AD + BC

[∵ AP + BP = AB

CR + DR = CD

AS + DS = AD

BQ + CQ = BC]

Hence, the right option is AB + CD = AD + BC

Given:

∠APB = 50°

Property 1:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 3:Sum of all angles of a triangle = 180°.

By property 1,

AP = BP (tangent from P)

Therefore, ∠PAB = ∠PBA

Now,

By property 3 in ∆PAB,

∠PAB + ∠PBA + ∠APB = 180°

⇒ ∠PAB + ∠PBA = 180° – ∠APB

⇒ ∠PAB + ∠PBA = 180° – 50°

⇒ ∠PAB + ∠PBA = 130°

By property 2,

∠PAO = 90°

Now,

∠PAO = ∠PAB + ∠OAB

⇒ ∠OAB = ∠PAO – ∠PAB

⇒ ∠OAB = 90° – 65° = 25°

Hence, ∠OAB = 25°

Given:

OA = 6 cm

OB = 8 cm

Property:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

By above property, ∆AOB is right-angled at ∠OAB (i.e., ∠OAB = 90°).

Therefore by Pythagoras theorem,

OA² + AB² = OB²

⇒ AB² = OB² – OA²

⇒ AB² = 8² – 6²

⇒ AB² = 64– 36

⇒ AB² = 28

⇒ AB= √28

⇒ AB= 2√7

Hence, length of tangent is 2√7 cm.

Given:AB and CD are two common tangents to circles which touch each other at C. If D lies on AB such that CD = 4 cm
To find: length of AB
Solution:

Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By the above property,

AD = BD = CD = 4 cm (tangent from D)

Now clearly,

AB = AD + BD

⇒ AB = AD + BD

⇒ AB = 4 cm + 4 cm

⇒ AB = 8 cm

Hence, AB = 8 cm

Property:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By the above property,

AE = AD (tangent from A)

AB = AC (tangent from A)

CD = CF (tangent from C)

BF = BE (tangent from B)

Now adding the above equations,

AB + BC + CA = AB + BF + FC + CA

⇒ AB + BC + CA = AB + BE + CD + CA

⇒ AB + BC + CA = AE + AD [∵ AE = AB + BE and AD = AC + CD]

⇒ AB + BC + CA = AD + AD [∵ AD = AE]

⇒ AB + BC + CA = 2AD

Hence, 2AD = AB + BC + CA

Given:

SQ = 6 cm

QR = 4 cm

Property:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

By above property, ∆ROQ is right-angled at ∠OQR (i.e., ∠OQR = 90°).

Diameter QS = 6 cm

⇒Radius (OQ) = 3 cm

Now by Pythagoras theorem,

OR² = OQ² + QR²

⇒ OR² = 3² + 4²

⇒ OR² = 9+ 16

⇒ OR² = 25

⇒ OR= √25

⇒ OR= 5 cm

Hence, OR= 5 cm.

Given:

AQ = 4 cm

BR = 6 cm

PC = 5 cm

Property:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By the above property,

AR = AQ = 4 cm (tangent from A)

BR = BP = 6 cm (tangent from B)

CP = CQ = 5 cm (tangent from C)

Now,

Perimeter of ∆ABC = AB + BC + CA

⇒ Perimeter of ∆ABC = AR + RB + BP + PC + CQ + QA

[∵ AB = AR + RB

BC = BP + PC

CA = CQ + QA]

⇒ Perimeter of ∆ABC = 4 cm + 6 cm + 6 cm + 5 cm + 5 cm + 4 cm

⇒ Perimeter of ∆ABC = 30 cm

Hence, Perimeter of ∆ABC = 30 cm

Given:

OP = 4 cm

∠OPA = 30°

Property:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

By above property, ∆POA is right-angled at ∠OAP (i.e., ∠OAP = 90°).

Now we know that,

Therefore,

⇒ AP = 2√3 cm

Hence, AP = 2√3 cm

Given:

Radius = 9 cm

OA = 15 cm

Property 1:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By the above property,

AP = AQ (tangent from A)

Property 2:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

By above property, ∆POA is right-angled at ∠OAP (i.e., ∠OPA = 90°).

Therefore by Pythagoras theorem,

AP² + PO² = AO²

⇒ AP² = AO² – PO²

⇒ AP² = 15² – 9²

⇒ AP² = 225 – 81

⇒ AP² = 144

⇒ AP = √144

⇒ AP = 12

AP + AQ = 12 cm + 12 cm = 24 cm

Hence, AP + AQ = 24 cm

Given:

Radius = OP = 5 cm

Distance of AB and XY = 8 cm

∵ Distance of AB and XY = 8 cm

And AB is parallel to XY

∴ PR = 8 cm

Join OB

Now,

OB = OP = 5 cm [radius]

Also,

OR = PR – PO

⇒ OR = 8 cm – 5 cm

⇒ OR = 3 cm

∴ By Pythagoras theorem in ∆ORB,

OB² = OR² + RB²

⇒ 5² = 3² + RB²

⇒ RB² = 5² – 3²

⇒ RB² = 25 – 9

⇒ RB² = 16

⇒ RB = 4

Now,

AB = AR + RB

⇒ AB = 2RB

⇒ AB = 2 × 4

⇒ AB = 8 cm

Hence, Length of chord = 8 cm

Property 1:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2:Sum of all angles of a triangle = 180°

By property 1, ∆PTO is right-angled at ∠OTP (i.e., ∠OTP = 90°).

By property 2,

∠OTP + ∠POT + ∠TPO = 180°

⇒ 90° + ∠POT + ∠TPO = 180°

⇒ ∠POT + ∠TPO = 180° - 90°

⇒ ∠POT + ∠TPO = 90°

Hence, ∠POT + ∠TPO = 90°

Given:

AB = 12 cm

BC = 8 cm

AC = 10 cm

Property:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By the above property,

AD = AF (tangent from A)

BD = BE (tangent from B)

CF = CE (tangent from C)

Clearly,

AB = AD + DB = 12 cm

BC = BE + EC = 8 cm

AC = AF + FC = 10 cm

Now,

AB – BC = 12 cm – 8 cm

⇒ (AD + DB) – (BE + EC) = 12 cm – 8 cm

⇒ AD + DB – BE – EC = 12 cm – 8 cm

⇒ AD + BE – BE – CF = 12 cm – 8 cm [∵ DB = BE and CF = CE]

⇒ AD – CF = 12 cm – 8 cm

⇒ AD – (10 cm – AF) = 12 cm – 8 cm [∵AF + FC = 10 cm ⇒ FC = 10 cm – AF]

⇒ AD – (10 cm – AF) = 4 cm

⇒ AD – 10 cm + AF = 4 cm

⇒ AD + AD = 4 cm + 10 cm [∵ AD = AF]

⇒ 2AD = 14 cm

⇒ AD = 7 cm

Hence, AD = 7 cm

Given:

AP = PB

Property:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By the above property,

AP = AQ (tangent from A)

BR = BP (tangent from B)

CQ = CR (tangent from C)

Clearly,

AP = BP = BR

AQ = AP = BR

Now,

AQ + QC = BR + RC

⇒ AC = BC [∵AC = AQ + QC and BC = BR + RC]

Hence, AC = BC

Given:

AP = 10 cm

OA = 6 cm

OB = 3 cm

Property :The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

By above property, ∆PAO is right-angled at ∠PAO (i.e., ∠PAO = 90°) and ∆PBO is right-angled at ∠PBO (i.e., ∠PBO = 90°).

Therefore by Pythagoras theorem in ∆PAO,

OP² = OA² + AP²

⇒ OP² = 6² + 10²

⇒ OP² = 36 + 100

⇒ OP= √136

Now by Pythagoras theorem in ∆PBO,

OP² = OB² + BP²

BP² = OP² – OB²

⇒ BP² = (√136) ² – 3²

⇒ BP² = 136 – 9

⇒ BP= √127

Hence, BP= √127 cm

Given:

∠RPQ = 60°

Property 1:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2:Sum of all angles of a triangle = 180°.

By property 1, ∆OPR is right-angled at ∠OPR (i.e., ∠OPR = 90°).

OP = OQ [∵ radius of circle]

∴ ∠OPQ = ∠OQP = 30°

Now by property 2,

∠OPQ + ∠OQP + ∠POQ = 180°

⇒ 30° + 30° + ∠POQ = 180°

⇒ 60° + ∠POQ = 180°

⇒ ∠POQ = 180° - 60°

⇒ ∠POQ = 120°

Hence, ⇒ ∠POQ = 120°

Property:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By the above property,

PD = PA (tangent from P)

QB = QA (tangent from Q)

RC = RB (tangent from R)

SC = SD (tangent from S)

Now,

PD + QB = PA + QA

⇒ PD + QB = PQ [∵PQ = PA + QA]

Hence, PD + QB = PQ

Given:

QP = 4.5 cm

Property:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By the above property,

PQ = PT = PR = 4.5 cm (tangent from P)

Now,

QR = PQ + PR

QR = PQ + PQ [∵PQ = PR]

QR = 2PQ

QR = 2 × 4.5 cm

QR = 9 cm

Hence, QR = 9 cm

Given:

∠QPB =50°

Property 1:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2:Sum of all angles of a triangle = 180°.

By property 1, ∆OPB is right-angled at ∠OPB (i.e., ∠OPB = 90°).

∠OPQ = ∠OPB – ∠QPB

⇒ ∠OPQ = 90° – 50° = 40°

And,

∠OPQ = ∠OQP [∵ OP = OQ (radius of circle)]

Now by property 2,

∠OPQ + ∠OQP + ∠POQ = 180°

⇒ 40° + 40° + ∠POQ = 180°

⇒ 80° + ∠POQ = 180°

⇒ ∠POQ = 180° - 80°

⇒ ∠POQ = 100°

Hence, ⇒ ∠POQ = 100°

Given:

APB = 30°

Property 1:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2:Sum of all angles of a triangle = 180°

By property 1,

PA = PB (tangent from P)

And,

∠PAB = ∠PBA [∵PA = PB]

By property 2,

∠PAB + ∠PBA + ∠APB = 180°

⇒ ∠PAB + ∠PBA + 30° = 180°

⇒ ∠PAB + ∠PBA = 180° - 30°

⇒ ∠ PAB + ∠ PBA = 150°

⇒ ∠ PBA + ∠ PBA = 150° [∵∠PAB = ∠PBA]

⇒ 2∠PBA = 150°

⇒ ∠PBA = 75°

Now,

∠PBA = ∠CAB = 75° [Alternate angles]

∠PBA = ∠ACB = 75° [Alternate segment theorem]

Again by property 2,

∠CAB + ∠ACB + ∠CBA = 180°

⇒ 75° + 75° + ∠CBA = 180°

⇒ 150° + ∠CBA = 180°

⇒ ∠CBA = 180° - 150°

⇒ ∠CBA = 30°

Hence, ∠CBA = 30°

Given:

QP = 4 cm

OQ = 3 cm

SR = 12 cm

SO’ = 5 cm

Property:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

By above property, ∆OPQ is right-angled at ∠OQP (i.e., ∠OQP = 90°) and ∆O’SR is right-angled at ∠O’SR (i.e., ∠O’SR = 90°).

By Pythagoras theorem in ∆OPQ,

OP² = QP² + OQ²

⇒ OP² = 4² + 3²

⇒ OP² = 16 + 9

⇒ OP² = 25

⇒ OP = √25

⇒ OP = 5 cm

By Pythagoras theorem in ∆O’SR,

O’R² = SR² + O’S²

⇒ O’R² = 12² + 5²

⇒ O’R² = 144 + 25

⇒ O’R² = 169

⇒ O’R = √169

⇒ O’R² = 13 cm

Now,

PR = PO + ON + NO’ + O’R

⇒ PR = 5 cm + 3 cm + 5 cm + 13 cm

⇒ PR = 26 cm

Hence, PR = 26 cm

Given:

AO’ = r

O’P = r

PO = r

AO = AO’ + O’P + PO

⇒ AO = r + r + r

⇒ AO = 3r

Now,

Hence, AO: AO’ = 3

Given:

OA = 5 cm

OQ = 3 cm

Property 1:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By property 1, ∆OAQ is right-angled at ∠OQA (i.e., ∠OQA = 90°).

By Pythagoras theorem in ∆OAQ,

OA² = QA² + OQ²

⇒ QA² = OA² – OQ²

⇒ QA² = 5² – 3²

⇒ QA² = 25² – 9²

⇒ QA² = 16

⇒ QA = √16

⇒ QA = 4 cm

By property 2,

BQ = BP (tangent from B)

And,

AQ = BQ = 4 cm [∵ Q is midpoint of AB]

PB = PC = 4 cm [∵ P is midpoint of BC]

Now,

BC = BP + PC

⇒ BC = 4 cm + 4 cm

⇒ BC = 8 cm

Hence, BC = 8 cm

Given:

PR = 7.5 cm

Property 1:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By property 1, ∆OSQ is right-angled at ∠OQS (i.e., ∠OQS = 90°) and ∆OPQ is right-angled at ∠OQP (i.e., ∠OQP = 90°).

∴ OQ ⊥ PS

∵ PO = OS [radius of circle]

∴ ∆POS is an isosceles triangle

Now,

∵ ∆POS is an isosceles triangle and OQ is perpendicular to its base

∴ OQ bisects PS

i.e., PQ = QS

By property 2,

PR = PQ = 7.5 cm (tangent from P)

Now,

PS = PQ + QS

⇒ PS = PQ + PQ [∵ PQ = QS]

⇒ PS = 7.5 cm + 7.5 cm

⇒ PS = 15 cm

Hence, PS = 15 cm

Given:

AB = 8 cm

PE = 3 cm

Property:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By above property,

AB = AC = 8 cm (tangent from A)

PE = CE = 3 cm (tangent from E)

Now,

AE = AC – CE

⇒ AE = 8 cm – 3 cm

⇒ AE = 5 cm

Hence, AE = 5 cm

Given:

_∠OPQ = 35°

Property 1:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 3:The sum of all angles of a triangle = 180°.

By property 1,

QP = QR (tangent from Q)

By property 2, ∆OPQ is right-angled at ∠OQP (i.e., ∠OQP = 90°) and ∆ORP is right-angled at ∠ORP (i.e., ∠ORP = 90°).

∴ OQ ⊥ QP

OR⊥ RP

Now,

∠OQP = ∠ORP = 90° [Property 1]

QP = QR [Property 2]

OP = OP [Given]

∴ ∆OPQ ≅ ∆OPR By SAS

Hence, ∠OPQ = ∠OPR = 35° By CPCTC

i.e. ∠a = 35°

By property 3,

∠OQP + ∠OPQ + ∠QOP = 180°

⇒ 90° + 35° + ∠QOP = 180°

⇒ 125° + ∠QOP = 180°

⇒ ∠QOP = 180° - 125°

⇒ ∠QOP = 55°

i.e. ∠b = 55°

Hence, ∠a = 35° and ∠b = 55°

Given:

_∠TQP = 60°

Property 1:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

By property 1,

TP = TQ (tangent from T)

⇒ ∠TPQ = ∠TQP = 60°

By property 2, ∆OPT is right-angled at ∠OPT (i.e., ∠OPT = 90°) and ∆OQT is right-angled at ∠OQT (i.e., ∠OQT = 90°).

Now,

∠OPQ = ∠OPT – ∠TPQ

⇒ ∠OPQ = 90° – 60°

⇒ ∠OPQ = 30°

Hence, ∠OPQ = 30°

Given:

PA = 4 cm

BP = 3 cm

AC = 11cm

Property:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By above property,

AP = AR = 4 cm (tangent from A)

BP = BQ = 3 cm (tangent from B)

QC = RC (tangent from C)

Clearly,

RC = AC – AR

⇒ RC = 11 cm – 4 cm

⇒ RC = 7 cm

Now,

BC = BQ + QC

⇒ BC = BQ + RC [∵ QC = RC]

⇒ BC = 3 cm + 7 cm

⇒ BC = 10 cm

Hence, BC = 10 cm

Given:

EK = 9 cm

Property:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By above property,

EM = EK = 9 cm (tangent from E)

DK = DH (tangent from D)

FM = FH (tangent from F)

Now,

Perimeter of ∆EDF = ED + DF + FE

⇒ Perimeter of ∆EDF = (EK – KD) + (DH + HF) + (EM – MF)

[∵ED = EK – KD

DF = DH + HF

FE = EM – MF]

⇒ Perimeter of ∆EDF = EK – KD + DH + HF + EM – MF

⇒ Perimeter of ∆EDF = EK – DH + DH + HF + EM – HF [∵DK = DH and FM = FH]

⇒ Perimeter of ∆EDF = EK + EM

⇒ Perimeter of ∆EDF = 9 cm + 9 cm

⇒ Perimeter of ∆EDF = 18 cm

Hence, Perimeter of ∆EDF = 18 cm

Given:

DE = 5 cm

DE DF

Join AE and AF

Property 1:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 3:Sum of all angles of a quadrilateral = 360°.

By property 1,

EF = ED = 5 cm(tangent from E)

And,

AE = AF [radius]

By property 2, ∠AED = 90° and ∠AFD = 90°.

Also,

∠EDF = 90° [∵ ED⊥EF]

By property 3,

∠AED + ∠AFD + ∠EDF + ∠EAF = 360°

⇒ 90° + 90° + 90° + ∠EAF = 360°

⇒ ∠EAF = 360° - (90° + 90° + 90°)

⇒ ∠EAF = 360° - 270°

⇒ ∠EAF = 90°

∵ All angles are equal and adjacent sides are equal ∴ AEDF is a square.

Hence, all sides are equal

⇒ AE = AF = ED = EF = 5 cm

Hence, Radius of circle = 5 cm

Given:

AB = 29 cm

AD = 23 cm

_∠B = 90°

DS = 5 cm

Property 1:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 3:Sum of all angles of a quadrilateral = 360°.

By property 1,

BP = BQ (tangent from B)

DS = DR = 5 cm (tangent from D)

AR = AQ (tangent from A)

Also,

OQ = OP (radius)

By property 2, ∆OQB is right-angled at ∠OQB (i.e., ∠OQB = 90°) and ∆OPB is right-angled at ∠OPB (i.e., ∠OPB = 90°).

Now by property 3,

∠PBC + ∠BQO + ∠QOP + ∠OPB = 360°

⇒ 90° + 90° + ∠QOP + 90°= 360°

⇒ 270° + ∠QOP = 360°

⇒ ∠QOP = 360° - 270°

⇒ ∠QOP = 90°

∵ adjacent sides (i.e., BP = BQ and OQ = OP) are equal and all angles are 90°

∴ quadrilateral OPBQ is a square

Now,

AD = 23 cm

⇒ AR + RD = 23 cm [∵ AD = AR + RD]

⇒ AR + 5 cm = 23 cm

⇒ AR = 23 cm – 5 cm

⇒ AR = 18 cm

⇒ AQ = AR = 18 cm

Now,

AB = 29 cm

⇒ AQ + QB = 29 cm [∵ AD = AR + RD]

⇒ 18 cm + QB = 29 cm

⇒ QB = 29 cm – 18 cm

⇒ QB = 11 cm

∵ OPBQ is a square

∴ OP = BQ = 11 cm

Hence, radius = 11 cm

Given:

BC = 12 cm

AB = 5 cm

Property 1:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 3:Sum of all angles of a quadrilateral = 360°.

By property 1,

AP = AQ (Tangent from A)

BP = BR (Tangent from B)

CR = CQ (Tangent from C)

∵ ABC is a right-angled triangle,
∴ by Pythagoras Theorem

AC² = AB² + BC²

⇒ AC² = 5² + 12²

⇒ AC² = 25 + 144

⇒ AC² = 169

⇒ AC = √169

⇒ AC = 13 cm

Clearly,

AQ + QC = AC
= 13 cm

⇒ AP + RC = 13 cm [∵ AQ = AP and QC = RC]

Also,

AB + BC = 5 cm + 12 cm
= 17 cm

⇒ AP + PB + BR + RC = 17 cm
[∵ AB = AP + PB and BC = BR + RC]

⇒ AP + RC + PB + BR = 17 cm

⇒ 13 cm + BR + BR = 17 cm
[∵ AP + RC = 10 cm and PB = BR]

⇒ 13 cm + 2BR = 17 cm

⇒ 2BR = 17 cm – 13 cm
= 4 cm

⇒ BR = 2 cm

Now,

∠BPO = 90° [By property 2]

∠BRO = 90° [By property 2]

∠PBM = 90° [Given]

Now by property 3,

∠BPO + ∠BRO + ∠PBM + ∠ROP = 360°

⇒ ∠ROP = 360° - (∠BPO + ∠BRO + ∠PBM)

⇒ ∠ROP = 360° - (90° + 90° + 90°)

⇒ ∠ROP = 360° - 270°

⇒ ∠ROP = 90°

Now, ∵ ∠ROP = 90° and BP = BR which are adjacent sides

∴ Quadrilateral PBRO is a square

⇒ PO = BR = 2 cm

Hence, Radius = 2 cm

Draw a tangent from a point T on B to P.

Property 1:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2:Sum of all angles of a triangle = 180°.

By property 1,

TA = TP (tangent from T)

TB = TP (tangent from T)

Now in ∆ATP,

TA = TP

∴ ∠APT = ∠PAT

And in ∆BTP,

TB = TP

∴ ∠BPT = ∠PBT

By property 2,

∠APB + ∠PBA + ∠PAB = 180°

⇒ ∠APB + ∠PBT + ∠PAT = 180°

⇒ ∠APB + ∠BPT + ∠ APT = 180° [∵ ∠APT = ∠PAT and ∠BPT = ∠PBT]

⇒ ∠APB + ∠APB = 180° [∵∠APB = ∠BPT + ∠APT]

⇒ 2∠APB = 180°

⇒ ∠APB = 90°

Hence, ∠APB = 90°

Given:

_∠QPR = 46°

Property 1:The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2:Sum of all angles of a quadrilateral = 360°.

By property 1, ∆OQP is right-angled at ∠OQP (i.e., ∠OQP = 90°) and ∆ORP is right-angled at ∠ORP (i.e., ∠ORP = 90°).

Now by property 2,

∠OQP + ∠ORP + ∠QOR + ∠QPR = 360°

⇒ ∠QOR = 360° - (∠OQP + ∠ORP + ∠QPR)

⇒ ∠ROP = 360° - (90° + 90° + 46°)

⇒ ∠ROP = 360° - 226°

⇒ ∠ROP = 134°

Hence, ∠ROP = 134°

Given:

PT = 3.8 cm

Property 1:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By property 1,

PQ = PT (Tangent from P)

PR = PT (Tangent from P)

Now,

QR = PQ + PR

⇒ QR = PT + PT

⇒ QR = 3.8 cm + 3.8 cm

⇒ QR = 7.6 cm

Hence, QR = 7.6 cm

Given:

AB = x cm

BC = 7 cm

CR = 3 cm

AS =5 cm

Property:If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By the above property,

AP = AS (tangent from A)

BP = BQ (tangent from B)

CR = CQ (tangent from C)

DR = DS (tangent from D)

Clearly,

QB = CB – CQ

⇒ QB = CB – CR [∵ CQ = CR]

⇒ QB = 7 cm – 3 cm

⇒ QB = 4 cm

Now,

AB = AP + PB

⇒ AB = AS + QB

⇒ AB = 5 cm + 4 cm

⇒ AB = 9 cm

⇒ AB = x = 9 cm

Hence, x = 9 cm